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Home , Homomorphism, Local ring, Maximal ideal, Module homomorphism, Zero divisor

Localization

Let A be a commutative . By a multiplicative S ⊂ A, we mean a such that 1 ∈ S −1 −1 and if s, t ∈ S then st ∈ S. We will construct a ring S A and a iS : A → S A that has the following : −1 For all s ∈ S, iS(s) is invertible in S A, and for every φ : A → B such −1 that φ(s) is invertible in B for all s ∈ S, there exists a unique homomorphism φS : S A → B such −1 that φ = φS ◦ iS : A → S A → B. In the definition of multiplicative set, 0 ∈ S is technically allowed. However, it is very obvious in that case that S−1A = (0), the , since 0 invertible implies 1 = 0. This is not very interesting. If 1 6= 0 in A, that is, if A 6= (0), and if 0 6∈ S, then S−1A 6= (0) will follow as well. If A is an integral and 0 6∈ S, then S−1A is a rather obvious subring of the field of fractions of A, namely, the subring of fractions whose denominators are elements of S. But even in this case the main point is not just the construction of S−1A, but rather the connections between ideals in A and S−1A and the many applications in commutative , number theory, and . The usual argument with universal properties shows that if S−1A exists, it is unique up to . The ring S−1A is called a localization of A. This terminology arises from consideration of rings of continuous functions, with values, say, in a field. If A is a ring of functions on a space X and if Y ⊂ X is a subspace, which could be a single point, let S = S(Y ) ⊂ A denote the subset of functions which have no zeros on Y . It is clear that S is a multiplicative set. The quotients a/s, with a ∈ A and s ∈ S, represent well-defined germs of functions near Y . That is, each such quotient defines a in some neighborhood of Y , specifically, the neighborhood where s is non-zero. These germs can be added and multiplied and form a ring, whose algebraic properties reflect the properties of X locally, that is, near Y . The ring A might have zero divisors, some elements of S might even be zero divisors, but the ring of germs of functions near Y which can be expressed as quotients a/s, still makes sense. It is perhaps useful to keep this picture in mind as motivation for the general construction below. Before constructing S−1A, we list some examples of multiplicative sets. (i) S is the set of all non-zero divisors in A. If A is an , this is just the set of all non-zero elements of A, and S−1A will turn out to be the field of fractions of A. (ii) S = A − P , where P ⊂ A is a prime . In this situation, the localization S−1A is often denoted A(P ), and referred to as the localization of A at the prime P . The ring A(P ) is a , that is, a ring with a unique . If A = A(V ) is the affine coordinate ring of a variety over an algebraically closed field K, and if P = I(p) ⊂ A is the maximal ideal corresponding to a point p ∈ V , then the localization A(P ) is a ring of germs of K-valued functions defined on open neighborhoods of p ∈ V in the . (It is a little harder to interpret the localization A(P ) as a ring of functions if P = I(W ) ⊂ A is a corresponding to an irreducible subvariety W ⊂ V .) (iii) It is obvious that arbitrary intersections of multiplicative sets are multiplicative sets. For example, if {Pj} is a family of prime ideals in A, then ∩j(A−Pj) = A−∪jPj is a multiplicative set. (iv) If s ∈ A is an element which is not , then S = {sn | n ≥ 0} is a multiplicative set. Suppose φ : A → B is a ring homomorphism so that some element s ∈ S is a zero divisor in A and φ(s) ∈ B is invertible. If as = 0 ∈ A then obviously φ(a) = 0 ∈ B, since φ(a)φ(s) = 0. This −1 observation makes it clear that iS : A → S A will not be injective if any elements of S are zero- divisors in A. A consequence is that the construction of S−1A is somewhat more subtle than, say,

1 the construction of the field of fractions of an integral domain. On the other hand, this observation about consequences of the existence of zero divisors in S motivates the precise definition of S−1A in general, which we now give. S−1A = {[a/s] | a ∈ A, s ∈ S}/∼, where [a0/s0] ∼ [a00/s00] if and only if there exists s ∈ S with a0s00s = a00s0s ∈ A. −1 The homomorphism iS : A → S A will be defined by iS(a) = [a/1]. The symbol [a/s] will −1 −1 0 0 0 00 0 00 0 00 0 00 name the element iS(a)iS(s) ∈ S A. So we certainly want [a /s ] = [a s /s s ] = [a s s/s s s], and [a00/s00] = [a00s0/s00s0] = [a00s0s/s00s0s]. The usual definition of equality of fractions, a0/s0 = a00/s00 if a0s00 = a00s0, would be adequate if no element of S were a zero divisor. The additional factor of s in the definition above allows for the possibility that a0s00 − a00s0 6= 0, but a0s00s − a00s0s = 0 for some s ∈ S. Exercise 1 Define addition and multiplication in S−1A, and verify that these operations are well- defined. Acknowledge that you ought to also check all ring axioms, associative laws, distributive laws, elements, etc, but don’t bother with all that.

−1 Exercise 2 Verify the universal property, stated at the outset, for iS : A → S A.

−1 Exercise 3 Characterize the set of all elements t ∈ A such that iS(t) is invertible in S A. (This subset of A is also a multiplicative set, called the saturation of S in A. Localizing A with respect to the saturation of S yields the same ring as localizing with respect to S. This can be seen instantly, since the universal property of S−1A implies that this ring also has the required universal property to be the localization with respect to the saturation of S.)

Next, we investigate relations between localization and ideals. Suppose I ⊂ A is an ideal with I ∩ S = ∅. Let A∗ = A/I and let S∗ be the of S in A∗. Then S∗ is a multiplicative set in A∗. The following exercise expresses the commutativity of localization and residue ring formation. One can either divide by an ideal first and then localize, or localize first and then divide by an appropriate ideal. Exercise 4 (S∗)−1A∗ =∼ S−1A/IS−1A. (Although it is easy enough to establish one-to-one ring in both directions, it is more elegant to verify that the ring on the right has the universal property of the ring on the left, hence it is isomorphic to the ring on the left.)

If φ : A → B is any ring homomorphism there are operations known as contraction and extension which relate ideals in B and A. If J ⊂ B is an ideal, the contraction is J c = φ−1(J) ⊂ A, the inverse image of J in A. If I ⊂ A is an ideal, the extension is Ie = φ(I)B, the ideal generated by the image of I in B. Both the contraction and extension operations preserve inclusions, that is, J1 ⊂ J2 c c ce ec implies J1 ⊂ J2, and similarly for extension. There are also obvious inclusions J ⊂ J and I ⊂ I , from which follow trivially the relations J cec = J c and Iece = Ie. The contraction has a number of reasonable properties, but very little can be said in general about extension. For example, the following properties of contraction are all easy to check, while the analogues for extension are all false. (i) If J is prime, primary, or radical, then J c is, respectively, prime, primary, or radical. c c c c (ii) rad J = (rad J) and ∩(Ji ) = (∩Ji) . −1 In case the homomorphism is a localization iS : A → S A, one can say more about the contraction operation, and, in addition, the extension operation has many good properties.

−1 ce e Exercise 5 If JS ⊂ S A is an ideal, then JS = JS. If I ⊂ A is an ideal, then I = {[a/s] | a ∈ I, s ∈ S} ⊂ S−1A and Iec = {b ∈ A | sb ∈ I, for some s ∈ S}.

2 Exercise 6 Ie ⊂ S−1A is proper if and only if I ∩ S = ∅. If I ⊂ A is prime or primary and I ∩ S = ∅, then Ie ⊂ S−1A is, respectively, prime or primary. Moreover, in each of these cases, Iec = I ⊂ A.

e e −1 Exercise 7 If I ⊂ A is any ideal, then rad I = (rad I) ⊂ S A. If I1 and I2 are ideals in A, e e e −1 then I1 ∩ I2 = (I1 ∩ I2) ⊂ S A.

It follows from Exercises 5 and 6 that contraction and extension define inclusion preserving between the sets of all prime or primary ideals, respectively, in S−1A and the sets of prime or primary ideals in A which are disjoint from S. It also follows, from Exercises 6 and 7, that a of an ideal I ⊂ A determines, by extension, a primary decomposition of Ie ⊂ S−1A. In this extension, any of the primary components of I that are not disjoint from S will generate the ideal in S−1A, hence disappear from the primary decomposition of Ie. Another simple consequence of the correspondence between prime ideals of S−1A and prime ideals of A disjoint from S is that in the case S = A − P , where P is a prime ideal of A, the −1 e localization S A = A(P ) is a local ring. Namely, the unique maximal ideal is P = {[x/s] | x ∈ P, s ∈ S} ⊂ S−1A, since, obviously, P ⊂ A contains all prime ideals of A disjoint from A − P . It is e also easy enough to verify directly that A(P ) is a local ring, since any element not in P is clearly a unit in A(P ), as it can be written [t/s], with both t and s not in P . The localization construction is also very important for modules. If S ⊂ A is a multiplicative set and if M is an A-, we construct an S−1A-module, S−1M, and an A- −1 fS : M → S M, with the following universal property: −1 For every S A-module NS and every A-module homomorphism φ : M → NS, there exists a −1 −1 −1 unique S A-module homomorphism φS : S M → NS such that φ = φS ◦fS : M → S M → NS. −1 (We point out that S A modules are also A-modules, via the ring homomorphism iS : A → S−1A.) The definition of S−1M, as well as the sum operation and by elements of S−1A, and the proofs that these operations are well-defined, proceeds identically, symbol by symbol, with the construction of S−1A and the ring operations in S−1A. S−1M = {[m/s] | m ∈ M, s ∈ S}/∼, where [m0/s0] = [m00/s00] if and only if there exists s ∈ S with m0s00s = m00s0s. −1 The fS : M → S M is given by fS(m) = [m/1]. The symbol [m/s] is interpreted as −1 −1 [1/s]fS(m), in the S A-module structure on S M. If you want to strengthen those brain wiring connections, go through the definition of the sum and scalar operations in S−1M, the proof that they are well-defined, and the proof that S−1M has the universal property stated above. Of course, the universal property characterizes S−1M uniquely, up to isomorphism as an S−1A-module.

Exercise 8 Prove that localization is an exact of modules. That is, an A-module ho- momorphism M → M 0 induces an S−1A-module homomorphism S−1M → S−1M 0, with the usual functorial properties, and, if M → M 0 → M 00 is an exact of A-modules, then S−1M → S−1M 0 → S−1M 00 is an of S−1A-modules.

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